# Interval List Intersections

I dedicated a whole page to this problem.

This was asked at of me at an interview and I flubbed it. I attempted this problem a month after the interview and I still didn't fully get it, and made numerous mistakes which I want to analyze.

Here is the algorithm I came up with:

Here is the code that I wrote:

vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {

int pa = 0, pb = 0;

vector<vector<int>> out;
vector<int> last = {-1, -1};
vector<int> cur;

while(pa < A.size() || pb < B.size() )
{   // get next interval
// deal with pointers going out of bounds
if      ( pa >= A.size()  )         { cur = B[pb++];   }
else if ( pb >= B.size()  )         { cur = A[pa++];   }
else if ( A[pa][0] <= B[pb][0] )    { cur = A[pa++];   }
else                                { cur = B[pb++];   }

// check if next intersects with last
int min_end   = min(last[1], cur[1]);
int max_start = max(last[0], cur[0]);

if(min_end >= max_start)
{
out.push_back({max_start, min_end});
}
last = cur;
}
return out;
}

Here's a review of the original algorithm with the mistakes I made.

Algorithm:

Code:

vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {

int pa = 0, pb = 0;
vector<vector<int>> out;

// if any pointer goes out of bounds, we are done checking!
while( pa < A.size() && pb < B.size() )
{
int min_end   = min(A[pa][1], B[pb][1]);
int max_start = max(A[pa][0], B[pb][0]);

if(max_start <= min_end)
{
out.push_back({max_start,min_end});
}

// push forward the pointer to the element with the smallest end
if(A[pa][1] < B[pb][1])
pa++;
else
pb++;
}
return out;
}
• interval_list_intersections.txt